KURATOWSKI's THEOREM Contents 1. Introduction 1 2. Basic Graph

KURATOWSKI’S THEOREM

YIFAN XU

Abstract. This paper introduces basic concepts and theorems in graph theory, with a focus on planar graphs. On the foundation of the basics, we state and present a rigorous proof of Kuratowski’s theorem, a necessary and suﬃ- cient condition for planarity.

Contents 1. Introduction 1 2. Basic Graph Theory 1 3. Planar Graphs 3 4. Kuratowski’s Theorem 7 Acknowledgments 12 References 12

1. Introduction The planarity of a graph, whether a graph can be drawn on a plane in a way that no edges intersect, is an interesting property to investigate. With a few simple theorems it can be seen that K5 (see figure 1) and K3,3 (see figure 2) are nonplanar. Kuratowski pushes this nearly effortless observation into a powerful theorem exposing the sufficient and necessary condition of planarity. Simple as the theorem appears to be, to prove this we need a significant amount of preparations. In this paper, we start with basic graph theory and proceed into concepts and theorems related to planar graphs. In the last section we will give a proof of Kuratowski’s theorem, which in general corresponds with that in Graph Theory with Applica- tions (see [1] in the list of references) but provides more details and hopefully more clarity.

2. Basic Graph Theory We need several deﬁnitions as a start. Deﬁnition 2.1. A graph G is an ordered pair (V (G),E(G)), consisting of a nonempty set V (G) of vertices and a set E(G) of edges, each edge a two-element subset of V . Denote |V (G)|, the number of vertices in the vertex set, by ν and |E(G)| by . Note that (i) E(G) can be empty and (ii) an edge can link a vertex to itself.

Date: August 28, 2017. 1 2 YIFAN XU

Figure 1. a diagram of K5

Deﬁnition 2.2. An edge whose ends are identical is a loop. Otherwise, the edge is a link. A graph is simple if it has no loops and no two links have the same pair of unordered vertices.

Deﬁnition 2.3. A vertex is incident to an edge if the vertex is one of the ends of the edge; two vertices are adjacent to each other if they are connected by an edge. The degree of a vertex v, denoted by deg(v), is the number of edges incident to v, with a loop counted as two edges. We let δ := minv∈V (deg v) and ∆ := maxv∈V (deg v).

Deﬁnition 2.4. A walk W is a set of alternating vertices and edges, denoted by W =v0e1v1e2...ekvk where ei (i ∈ [1, k], i ∈ N) links vi−1 with vi. A walk where all ei’s are distinct is a trail. In addition, if all vertices are distinct, then W is a path. A graph G is connected if there exists a path between every pair of vertices in G. A walk is closed if it has positive length and identical ends. A closed trail with distinct vertices is a cycle. A family of walks is internally disjoint if no vertex is an internal vertex of more than one walk in the family.

Deﬁnition 2.5. H is a subgraph of G if V (H) ⊆ V (G), E(H) ⊆ E(G), and endpoints of all edges in E(H) are included in V (H). In addition, if H is a maximally connected subgraph, H is a component of G. The number of components in G is denoted by ω(G). An edge e is a cut edge if ω(G − e) > ω(G).

Deﬁnition 2.6. A complete graph is a graph where each pair of vertices is connected by a unique edge. We let Km denote the complete graph on m vertices. A graph G is bipartite if its vertex set V can be divided into two nonempty subsets X and Y such that every edge in G connects one vertex in X to another one in Y . A graph G is complete bipartite if for all x ∈ X, y ∈ Y , x is connected to y by a unique edge. When X contains m vertices and Y contains n vertices, G is denoted by Km,n.

Theorem 2.7. For a graph G with vertex set V and edge set E, X deg(v) = 2 v∈V KURATOWSKI’S THEOREM 3

X1 X2 X3 a ! caa #c !!# s c a # s c ! # s c a#a !c! # c# a!a! c# # c!! aa# c #!! c # aac !##!! cc## aacca Ys1 Ys2 Ys3

X1 Y1 T S s T s S T S T S Y3 T S X2 S T s S T s S T S TT X3S Y2 s s Figure 2. two diagrams of K3,3

Proof. For v ∈ V and e ∈ E, we let n(v) denote the set of edges incident to v and m(e) the set of endpoints of e. Then X X X deg(v) = 1 v∈V v∈V e∈n(v) X X = 1 e∈E v∈m(e) = 2. 3. Planar Graphs Definition 3.1. A way to draw the graph, representing vertices by points and edges by lines connecting points, is called a diagram of the graph. A diagram is embedded in the plane. A graph that has a diagram whose edges do not intersect anywhere besides their ends (i.e., vertices) is called a planar graph. The diagram is then called the planar embedding of a planar graph, or simply, a plane graph. Definition 3.2. Closures of regions partitioned by a plane graph are faces, the number of which is denoted by φ. In a plane graph, the degree of a face f, denoted by deg(f), is the number of edges incident to f, with cut edges being counted twice. Definition 3.3. A dual of a plane graph G, denoted by G∗, can be constructed as follows: every vertex v∗ in G∗ corresponds to a face f in G and every edge e∗ in G∗ corresponds to an edge e in G. Two vertices in G∗, v∗ and w∗, are joined by the edge e∗ if and only if their corresponding faces in G, f and g, are separated by e. 4 YIFAN XU

Theorem 3.4. For G a plane graph, let F (G) denote the set of faces in G. The following holds: X deg(f) = 2. f∈F (G) Proof. Consider the dual, G∗, of G. By Theorem 2.13, X deg(v∗) = 2∗. v∗∈V ∗ By deﬁnition of a dual graph, ∀f ∈ F (G), deg(f)=deg(v∗), and =∗.Then X X deg(f) = deg(v∗) = 2∗ = 2. f∈F (G) v∗∈V ∗ Theorem 3.5. (Euler’s Formula) For G a connected plane graph, the following relationship holds: ν − + φ = 2. Proof. We prove this by induction on . (1) Basic step: when = 0, since G is connected, G must be trivial. Then ν = 1, = 0, φ = 1. ν − + φ = 1 − 0 + 1 = 2. Clearly the formula holds. (2) Inductive step: suppose for G with (G) = k, ν(G) − (G) + φ(G) = 2. Let H be a planar graph such that G is a subgraph and (H) = k + 1. There are 3 cases: Case(a): e is a loop added on some v ∈ V . Then ν(H) = ν(G). Since H needs to remain a plane graph, the loop does not intersect with any other edge in G. Therefore the loop constitutes a new face, φ(H) = φ(G) + 1, which then gives us ν(H) − (H) + φ(H) = ν(G) − ((G) + 1) + (φ(G) + 1) = ν(G) − (G) + φ(G) = 2.

Case(b): e adds a link between two vertices v1, v2 in G. Since G is connected, there is a path between v1 and v2. The addition of e creates a new cycle, and hence a new face (if there are more than one path between v1 and v2, there is necessarily one that borders the new face). Therefore ν(H) = ν(G), φ(H) = φ(G) + 1, (H) = (G) + 1. Similar to (a), the formula holds. Case(c): e links some v1 ∈ V (G) to a new vertex v2. Then ν(H) = ν(G) + 1. Since v2 is a new vertex, e lies in some face bordering v1, without creating any new face. Then φ(H) = φ(G). In this case, ν(H) − (H) + φ(H) = (ν(G) + 1) − ((G) + 1) + φ(G) = ν(G) − (G) + φ(G) = 2. Therefore in all 3 cases, the formula holds in H. By the principle of induction, the formula holds for all connected plane graphs. KURATOWSKI’S THEOREM 5

Corollary 3.6. For G a simple planar graph with ν ≥ 3, ≤ 3ν − 6. Proof. Since G is a simple graph with ν(G) ≥ 3, the planar embedding of G, G0, is also simple with ν(G0) ≥ 3. For f ∈ F (G0), there exist 3 cases: (1) if f is bounded by a cycle, since G0 is simple, the minimum size of a cycle is 3 and degG0 (f)≥3; (2) if f is incident to some edges in addition to a cycle, degG0 (f)>3; (3) if f is incident to no cycle at all, then there exists at least two cut edges on the boundary of f—if not, the rest of the boundary must be a cycle, which contradicts the condition—which means degG0 (f) ≥ 4 > 3. Therefore, for every f ∈ F (G0),

X 0 degG0 (f) ≥ 3φ(G ) then by Theorem 3.7, X deg(f) = 2(G0) ≥ 3φ(G0) f∈F (G0) which yields 2 φ(G0) ≤ (G0). 3 Then by Theorem 3.8, 2 ν(G0) − (G0) + φ(G0) = 2 ≤ ν(G0) − (G0) + (G0) 3 1 = ν(G0) − (G0) 3 1 = ν(G) − (G). 3

Therefore, 1 ν(G) − (G) ≥ 2 3 and ≤ 3ν − 6.

Corollary 3.7. K5 is nonplanar. Proof. We have 5 (K ) = = 10 > 3ν(K ) − 6 = 9, 5 2 5 by Corollary 3.9, K5 is nonplanar.

Corollary 3.8. K3,3 is nonplanar. Proof. We are going to prove this by contradiction. Suppose there exists a planar embedding of K3,3. In a simple bipartite graph, the minimum size of a cycle is 4, which means for f ∈ F (K3,3), deg(f) ≥ 4 6 YIFAN XU

Figure 3. an example of 3 bridges on a cycle (the red one is skew to the blue one; the cyan one avoids the other two; red and cyan outer bridges, blue an inner bridge)

(the argument similar to that in Corollary 3.9). By Theorem 3.7, X 4φ ≤ deg(f) = 2 = 18 f∈F (G) then since φ ∈ N, φ ≤ 4. By Theorem 3.8, ν − + φ = 2 ≤ 6 − 9 + 4 = 1, which is impossible. Therefore, K3,3 is nonplanar. Definition 3.9. Let H be a subgraph of a graph G. Define an equivalence relation ∼ on E(G) \ E(H) as follows: a ∼ b if there is a walk W such that a and b are the first and last edge in W respectively, and that no internal vertex of W is in V (H). A bridge of H in G is a subgraph of G − E(H) induced by an equivalent class of ∼ (a bridge containing e is the subgraph containing every e0, e0 ∼ e, e and e0 edges of G). For a bridge B of H, we define vertices of attachment of B to H as the vertices in the set V (B) ∩ V (H). Let C be a cycle. Then two bridges of C, B1 and B2, are skew if two vertices of attachment of B1, say u1 and v1, and two of B2, u2 and v2, appear in the order of u1, u2, v1, v2 on C. Definition 3.10. Suppose C is a cycle in a planar embedding of a planar graph G. Then for some bridge B of C, B is contained entirely in either Int(C) (the region inside C) or Ext(C) (the region outside C). A bridge in Int(C) is an inner bridge, while one in Ext(C) is an outer bridge. In the planar embedding, inner (or outer) bridges avoid each other: for all B1,B2 two inner(outer) bridges, all vertices of attachment in B1 lie on the arc uv of C which contains no vertices of attachment of B2 other than u and v. KURATOWSKI’S THEOREM 7

Definition 3.11. In some planar embedding G1 of a planar graph G, an inner bridge B of C (a cycle in G) is transferrable if there exists another planar embedding G2 of G, where B is an outer bridge but everything else remains the same as in G1. Theorem 3.12. Let G be a plane graph and C a cycle in G. An inner bridge B of C is transferrable if B avoids every outer bridge of C. Proof. Find an inner bridge B that avoids every outer bridge. Then we can find a face in Ext(C) whose boundary contains all vertices of attachment of B. Drawing B on the new face gives us another planar embedding, which means B is transferrable. 4. Kuratowski’s Theorem In 1930, Kuratowski published the theorem giving a necessary and sufficient condition for planarity. Kuratowski’s Theorem states that a graph is planar if and only if it contains no subdivision of K5 or K3,3. To prove this theorem, we first need some simple lemmas. Lemma 4.1. Every subgraph of a planar graph is planar. Proof. If G is planar, then there exists a planar embedding of G. For every subgraph H of G, we can find the vertices and edges of H in the planar embedding of G. This is how we can construct a planar embedding of H. Definition 4.2. A subdivision of an edge is the operation where the edge is replaced by a path of length 2, the internal vertex added to the original graph. A subdivision of a graph G is a graph achieved by a sequence of edge-subdivisions on G. Lemma 4.3. Every subdivision of a nonplanar graph is nonplanar. Proof. Suppose for G, there exists a planar embedding of its subdivision, G0. When we remove the vertices created in edge-subdivisions, and reconstruct the original edge (without changing the shape and position of the path), we get a planar embedding of G and find G planar. Therefore, if G is nonplanar, every subdivision of G is nonplanar. From the two lemmas above the necessity easily follows. Then it suffices to prove that the condition is also sufficient. In order to show if G contains no subdivisions of K5 or K3,3, G is planar, it is equivalent to show that if G is nonplanar, G has to contain some subdivision of K5 or K3,3. Two definitions are necessary before we get to the strategy of proving the equivalent statement. Definition 4.4. For a graph G, a vertex cut V 0 is a subset of V whose removal renders G − V 0 disconnected (when we remove a vertex, we remove the vertex as well as all edges incident to it). The connectivity of G, denoted by κ(G), is the minimum size of the vertex cut V 0. A graph G is said to be k-connected if k ≤ κ(G).

Deﬁnition 4.5. For a graph G, H is a proper subgraph of G if V (H)(V (G) and E(H)(E(G). A minimal nonplanar graph is a nonplanar graph that does not have any nonplanar proper subgraph. Clearly, it suﬃces to prove the statement for all G some minimal nonplanar graph. The strategy is as follows: (1) Show that if minimal nonplanar graphs without any subdivision of K5 or K3,3 8 YIFAN XU as subgraphs did exist, they would be 3-connected and simple. (2) Show that every 3-connected graph with no subdivision of K5 or K3,3 as subgraphs is in fact planar. This is how we arrive at a contradiction, forcing the original statement to be true. To show (1), we need to establish a few more lemmas. Lemma 4.6. A minimal nonplanar graph is 2-connected.

Proof. First show that a minimal nonplanar graph is 1-connected (connected). Sup- pose G is disconnected and nonplanar, but all of its components are planar. With- out loss of generality, suppose G has two components, G1 and G2. Since G1 and G2 are both planar, we can add a planar embedding of G1 to one of the faces of a planar embedding of G2 (the infinite face for example), which yields a planar embedding of G, a contradiction. Then we show that it is 2-connected. Suppose G is nonplanar, with κ(G) = 1. By definition of connectivity, there exists a vertex v such that G−v is disconnected. Without loss of generality, suppose G − v has two components, H1 and H2. We know that H1 ∪ v and H2 ∪ v are both planar. In the planar embedding of each, we can find a face f whose boundary contains v. With stereographic projection, we can get a planar embedding for each of H1 ∪ v and H2 ∪ v where v lies on the boundary of the unbounded face, by placing the point at infinity on the sphere inside f. Then we can combine H1 ∪ v and H2 ∪ v by merging v and get a planar embedding of G, a contradiction. Therefore if G is a minimal nonplanar graph, G is 2-connected.

Lemma 4.7. If G is a graph that has the fewest edges possible among all connected nonplanar graphs with no subdivision of K5 or K3,3, then G is 3-connected. Proof. The hypothesis suggests G is a minimal nonplanar graph. By the previous lemma, G is 2-connected. Suppose κ(G) = 2. Then there exists a vertex cut {u, v} such that G−{u, v} is disconnected. Name the components of G−{u, v} H1, H2,..., Hk. Construct M1, M2,..., Mk, where Mi is Hi ∪ {u, v} with the addition of a new edge uv. We claim here that among Mi, 1 ≤ i ≤ k, there exists at least one Mi that is nonplanar. Below is a proof for the claim: Suppose all Mi’s are planar for 1 ≤ i ≤ k. Then there is a planar embedding for each. Since {u, v} and the edge uv are the only part Mi’s share, we can merge the planar embeddings of Mi’s and get a planar embedding of G + uv (G ∪ {uv}), which means G + uv is planar. Then by Lemma 4.1, G is planar, a contradiction. Therefore, there exists some Mj where 1 ≤ j ≤ k that is nonplanar. It is clear that (Mj) < (G). But since G, by the original hypothesis, is the smallest connected nonplanar graphs with no subdivision of K5 or K3,3, Mj must have some subdivision of K5 or K3,3. Moreover, since G contains no such subdivision, Mj is not a subgraph of G, which means G does not have an edge uv. Now we combine Mj − uv with Mp − uv where p 6= j, 1 ≤ p ≤ k by merging the vertices u and v and get a subgraph of G. Since Mp − uv is connected, there exists a path between u and v. When we combine this path with Mj −uv, we get a subdivision of K5 or K3,3. This means G contains such a subdivision, a contradiction. Therefore, G has to be 3-connected.

Now (1) has been shown. To complete (2), we again need some lemmas. KURATOWSKI’S THEOREM 9

Lemma 4.8. (Whitney’s Theorem) Let G be a graph with ν ≥ 3. Then G is 2- connected if and only if for all u, v ∈ V (G), there are at least two internally-disjoint paths between them. Proof. (⇐) If any two vertices in G are connected by at least two internally-disjoint paths, then clearly there exists no 1-vertex cut (since no matter which vertex is removed, between every two vertices that remain, there still exists at least one path between them). Hence G is 2-connected. (⇒) Suppose G is 2-connected. We shall prove this direction by induction. Take two vertices u, v ∈ V (G). Denote the number of edges in the shortest paths between them by d(u, v). (a) Basic step: when d(u, v) = 1. Since G is 2-connected, there exists another path connecting u, v, which does not contain the edge uv. (b) Inductive step: Suppose there exist at least two internally-disjoint paths for all u, v with d(u, v) ≤ k. For x, y with d(x, y) = k + 1, find a path P0 of length d(x,y) between x, y and a vertex z that is closest to y (d(y, z) = 1) in P0. Then d(x, z) = d(x, y)−1. By the inductive hypothesis, there exist two internally-disjoint paths P1 and P2 between x, z. Since G is 2-connected, there exists another path Q between x, y that does not contain z (or {w} would be a 1-vertex cut). Let w be the vertex in (Q ∩ (P1 ∪ P2)) that is closest to y on Q. Without loss of generality we suppose w is contained in P1. Then we can find two internally-disjoint paths between x, y: the first one would be the part from x to w on P1 combined with the part from w to y on Q; the second one would be P2 combined with the edge zy. Lemma 4.9. If G is simple and 3-connected and uv is an edge in G, then G − uv is 2-connected. Proof. We want to show that for all a, b ∈ V (G − uv), there exist at least two internally-disjoint paths between them. In other words, we want to show for every two vertices of G − uv, there exists a cycle they both lie on. We prove this by discussing 3 cases. (1) {a, b} = {u, v}. Clearly ν(G) ≥ 4. Pick another two vertices c and d in G − uv. Without loss of generality, assume u = a. Now consider u and c. Since G is 3-connected, G does not contain any 2-vertex cut, which means when v and d are removed, u and c are still connected. In other words, there exists a path P1 between u and c which does not contain v and d. Similarly, there exists a path P2 between c and v that avoids u and d, a P3 between v and d that avoids u and c, and finally a P4 between d and u that avoids c and v. But then u and v lie on the same cycle u-P1-c-P2-v-P3-d-P4-u. (2) One and only one of {a, b} is u or v. Without loss of generality, let a = u and b 6= v. Find c 6= b that is neither u nor v. Then following the similar argument as in (1), we can find a path P1 avoiding c and v between u and b, a P2 avoiding u and v between c and b, and a P3 avoiding v between c and u. Again u-P1-b-P2-c-P3-u is a cycle containing u, b. (3) Neither of {a, b} equals u or v. Once again following the same argument, we can find a path P1 avoiding u, v between a, b, a P2 avoiding u, a between v, b, and a P3 avoiding u, b between a, v. Then a-P1-b-P2-v-P3-a is a cycle containing a, b. Since in all 3 cases, we construct a cycle without the edge uv where both a, b lie, the same cycles can be constructed in G − uv, which means G − uv must be 2-connected. 10 YIFAN XU

s x1 s

y1 y2 s s

x2 s v s

s y2 l s l l x l x 1 l 2 l s l s l l y1 v s s Figure 4. Case (1) and (2), with colors indicating bipartition

Now we are ready for the actual proof. Theorem 4.10. (Kuratowski’s Theorem) A graph is planar if and only if it does not contain any subdivision of K5 or K3,3. Proof. (⇒) It is true by the ﬁrst two lemmas in this section. (⇐) Suppose there exists a nonplanar graph that does not contain any subdivision of K5 or K3,3. Without loss of generality, let G be a nonplanar graph that contains no subdivision K5 or K3,3 and has the fewest edges possible. Then G is a minimal nonplanar graph. By Lemma 4.7, G is 3-connected (and simple). Take two adjacent vertices u, v ∈ V (G). Consider the subgraph G − uv. By minimality, G − uv is planar. By Lemma 4.9, G − uv is 2-connected. By Lemma 4.8, there are at least two internally-disjoint paths between u and v. In other words, u and v lie on some common cycle. Among all cycles containing u and v in a planar embedding of G − uv, ﬁnd C0 with the most edges in Int(C0). Now consider the bridges of C0 in G − uv (if G − uv does not contain any bridge of C0, then it is clear that with the addition of edge uv, the graph is still planar, which means G is planar, a contradiction). Suppose there exists a bridge with only one vertex of attachment v1. Then v1 is a one-vertex cut of G − uv, which KURATOWSKI’S THEOREM 11

u(y3)

z x1(y1) x2(y2) s s s

v(y4) s

u(y3) ¡ ¡ s ¡ ¡ z ¡ z x1(y1) 1 ¡ 2 x2(y2) ¡ s s ¡ s s ¡ ¡ ¡ ¡ v(y4) s

Figure 5. Case (3) and (4)

contradicts the condition that G − uv is 2-connected. Therefore, all bridges of C0 in G − uv have at least two vertices of attachment. Moreover, if an outer bridge of C0 has more than 2 vertices of attachment, we can always find a new cycle that contains parts of the outer bridge and has more edges in its interior. Therefore all outer bridges of C0 have exactly 2 vertices of attachment. Following the same argument, if an outer bridge avoids the arc uv, then there would be another cycle with more edges in the interior. Hence all outer bridges overlap the arc uv, that is, for any outer bridge, not all vertices of attachment lie on the same arc uv. Also, if the size of an outer bridge is more than one, there exists a vertex that is not on C0 in the bridge. Then the two vertices of attachment form a 2-vertex cut of both G − uv and G, which contradicts the condition of G being 3-connected. Therefore, we can conclude that all outer bridges of C0 have 2 vertices of attachment, have the size 1, and overlap the arc uv. Find an outer bridge B1 and an inner bridge B2 that overlap. Justification for finding such B1,B2 is as follows. If all bridges of C0 are inner (outer) bridges, then we can draw the edge uv in the exterior (interior) of C0 and achieve a planar 12 YIFAN XU embedding of G, which contradicts the hypothesis. Hence C0 has to have both inner and outer bridges. The reason why there exists a pair that overlap is that if not, then every inner bridge of C0 avoids every outer bridge, and by Theorem 3.16, all inner bridges of C0 are transferrable. We can then find a planar embedding of G − uv where C0 have only outer bridges, which again contradicts the hypothesis. Let the vertices of attachment of B1 be x1, x2, and those of B2 be y1, y2, y3, .... We know that B2 overlaps the arc uv, and is skew to B1. We consider 4 cases in terms of the relative position of B1 and B2. Without loss of generality, we assume that u, x2, v, x1 lie on the cycle in a clockwise order.

(1) Among all vertices of attachment of B2, there exist y1, y2 such that y1 lies between x1 and v, y2 between u and x2. Then G contains a subdivision of K3,3, which is against our assumption. (2) There exist y1, y2 such that y1 lies between x2 and v, y2 between x1 and u. Still G contains a subdivision of K3,3, a contradiction. (3) There exist {y1, y2, y3, y4} = {x1, x2, u, v} such that the u-v path P1 and the x1-x2 path P2 have one and only one vertex z in common (P1 and P2 must have some vertices in common because of the planarity of G − uv). Then G contains a subdivision of K5, a contradiction. (4) There exist {y1, y2, y3, y4} = {x1, x2, u, v} such that P1 and P2 have more than one vertex in common. Then G again contains a subdivision of K3,3. By now we have covered every possible case and derived a contradiction from each of them. Therefore, the theorem is true. Acknowledgments. It is a pleasure to thank my mentor, Reid Harris, for his help- ful guidance and advice. I would also like to thank Professor Babai for introducing me to graph theory and Professor May for organizing the REU.

References [1] J. A. Bondy and U. S. R. Murty. Graph Theory with Applications. The Macmillan Press Ltd., 1982, page 1-15, 143-156. [2] Walter Klotz. ”A constructive proof of Kuratowski’s theorem.” https://www.researchgate.net/ publication/256078009.